Molar heat of vaporisation of a liquid is `6 kJ mol^(-1)`. If the entropy change is `16 J mol^(-1) K^(-1)` the boiling point of the liquid is

Options

A. A) `375^(@)C`
B. B) `375 K`
C. C) `273 K`
D. D) `102^(@)C`

Answer

b

Explanation

`/_\S = 16 J mol e^(-1) K^(-1)`
`T_(b.p.)=(/_\H_("vapour"))/(/_\S_("vapour")=((6xx1000)/(16))=375k`